small math problem

This post has been edited 2 time(s), it was last edited by tevere: 13.11.2009 20:12.

ok, so in no limit holdem the board on the river is 4spades and another ofsuit card no pairs. i hold the 6 of spades (the 4 spades on the board are higher than my 6) and there is me and 3 guys in this pot. check all the way to the river when somone makes a decent , not big bet.

what is the probablilty of a man holding a biger spade than me? i figure there are 4 cards that beat me out of 8 spades, but how do i calculate when i got 3 oponents (as in 6 active cards out of 45) what is the math to this.
thank you!

Ok firstly it is out of 7 spades, not 8. 13-6 = 7.
Think about it...
There are 6 chances that one of them has a higher spade, because there are 6 hole cards held by opponents that you have to worry about.

I'd work it out like this.
First card of first hand = 4/45
if no high spade then 2nd card of first hand = 4/44
if no high spade then 1st card of 2nd hand = 4/43
etc. 4/42
etc. 4/41
if no high spade then 2nd card of 3rd hand = 4/40
ADD (not multiply!!!) those all together and you get probability of at least 1 spade.

You could do that if you want a precise calculation, or just go 4/45 x 6.

the probability of some1 having higher spade then you isnt that important.. Ask yourself, what kind of hand would bet out after 4th spade comes up ?
do you expect some with 5 of spades to fire and expect 4s to call ?

Originally posted by GooRukYONG
the probability of some1 having higher spade then you isnt that important.. Ask yourself, what kind of hand would bet out after 4th spade comes up ?
do you expect some with 5 of spades to fire and expect 4s to call ?

LOL I agree to some extent but you would need to see the bet sizing and patterns. I've bet in situations similar when I don't have a spade in my hand, sometimes you win just by betting. Keep in mind he said decent bet. I would never call a huge bet or allin or anything though.

If you assume the other players have any 2 cards with equal probability given the betting patters (probably not a great assumption!) then the easiest way of finding the probability that someone has a higher spade is by finding the probability that nobody has a higher spade first.

This is (41/45)*(40/44)*(39/43)*(38/42)*(37/41)*(36/40) which is roughly 0.55,
so the probability someone DOES have a higher spade is 1 - 0.55 = 0.45.

There are 6 cards in your opponents' hands, each of which is one of the 45 remaining cards in the deck. The probability the first is not one of the 4 higher spades is (45-4)/45 = 41/45. The probability that neither the first or the second are higher spades is P(first is not a high spade)*P(second is not a high spade GIVEN that the first is not a high spade). If we know that the first is not a high spade, then the chance that the second is not a high spade is 40/44 (since one card is now accounted for by the first card). If you continue this process you get the formula above.

You should never be adding probabilities in situations like this...if you have a 1% chance of winning something and you try it twice, the chance that you win at least once is NOT 2%!!! It is 1-(1-.01)*(1-.01) = .0199. The difference here is negligible but if you have two coinflips, the probability you win one is not .5+.5 = 1 but .75 (even though it may seem like .35 sometimes
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